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Answer:
0.212 mm,
R
right end
=−
1
.
16 kN,
M
right end
=−
89
.
28 kN
·
m,
R
left end
=
982
.
14 kN
3.28 Determine the force in the middle spring of the beam of Fig. P3.24b with five springs.
Let
k
a
=
50 kN/m,
k
b
=
60 kN/m,
k
c
=
55 kN/m,
k
d
=
100 kN/m,
k
e
=
k
n
=
10
9000 cm
4
,
E
84 GN/m
2
,
a
1
kN/m,
L
=
2
.
5m,
p
0
=
5 kN/m,
I
=
=
75
.
=
0
.
5m,
a
2
=
=
.
=
1m,
a
3
1
5m,and
a
4
2m.
Answer:
1.307 kN
3.29 Find the redundant f
or
ce
R
between two simply supported beams crossing each other
and loaded by force
P
(Fig. P3.29).
FIGURE P3.29
Crossed beams.
Answer:
R
=
8
P
/
9
3.30 A thin circular ring is subjected to equal and opposite loads
P
as shown in Fig. P3.30a.
Determine the moments in the ring. Consider only bending in the bar.
Choose
M
∗
of Fig. P3.30b as the redundant force. Use
∂
U
i
/∂
M
∗
=
Hint:
0tofind
M
∗
. Because of symmetry,
2
π/
2
0
M
2
R
2
EI
(
−
α)
PR
1
cos
U
i
=
M
∗
−
α
=
d
,
M
2
PR
2
Answer:
M
=
0
.
182
PR
−
(
1
−
cos
α)
,M
max
=−
0
.
318
PR
under the load.
3.31 Find the vertical displacement in the ring of Fig. P3.30a at the position of loading.
Hint:
Since the internal moments in the ring are known from the previous prob-
lem, a direct applic
ati
on of Castigliano's theorem, part II will yield the displace-
ments 2
V
P
=
∂
U
i
/∂
P
.
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